"Big O" notation crops up everywhere in analytic number theory. Roughly speaking, we say $f(x) = O(g(x)$ if there exists a positive constant $M$ s.t. $\lvert f(x) \rvert \leq M \lvert g(x) \rvert $ for sufficiently large $x$ (we can also define it near other points, such as when $x$ is near $0$, but here, I'm assuming that $x$ is going off to infinity).

Another notation I've met recently is $f(x) \ll g(x)$. In papers, authors seems to use "big O" and double "less than" signs interchangeably.

Are these notations equivalent, or do they have slightly different meanings?

  • $\begingroup$ I would say that $\ll$ is more along the lines of "(aymptotically) much less than," i.e. I would take $f(x) \ll g(x)$" as either $f(x) = o(g(x))$ or "$f(x) \ll c g(x)$ for a small enough constant $c\in(0,1)$." But in the end, it's a matter of notations (if the authors don't (re)define the notation, however, I'd would read it as this.) $\endgroup$ – Clement C. Nov 13 '15 at 12:06
  • $\begingroup$ As a rule, it is much easier to answer questions about specifics. Definitions vary. Name a book, quote a passage. Big-O is fairly normal (although the meaning varies for computer science.) Double-less-than is less often used formally. $\endgroup$ – Thomas Andrews Nov 13 '15 at 12:29
  • $\begingroup$ As others have said it depends on context. In differential equations, $f(x) \ll g(x)$ would often mean $f(x) = o(g(x))$, but in number theory it would often mean $f(x) = O(g(x))$. $\endgroup$ – Antonio Vargas Nov 13 '15 at 13:23
  • $\begingroup$ @ThomasAndrews, is it pronounced "double less than" or "much less than"? $\endgroup$ – user231343 Oct 2 '18 at 10:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.